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Let $G$ be a compact Lie group and $a\in\mathfrak{g}^*$ (dual of Lie algebra of Lie group $G$). Then let $\mathcal O_a$ be a coadjoint orbit. Then every co-adjoint orbit is Kähler manifold and also projective variety. How can we compute the Kodaira dimension of co-adjoint orbit as projective variety?

Motivation: The Kodaira dimension of co-adjoint orbits are important, because we can classify these type of projective varieties by Kodaira dimension which is birationally invariant.

In fact I am looking for

$$\kappa(\mathcal O_a)=\limsup_{m\to \infty}\frac{\log\text{dim}H^0(\mathcal O_a, K_{\mathcal O_a}^{\otimes m})}{\log m}$$

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  • $\begingroup$ In fact, coadjoint orbit is hyper-kahler variety, so normally the canonical bundle is trivial and Kodaira dimension must be 0. Am I wrong? $\endgroup$
    – user21574
    Commented Oct 29, 2014 at 13:19
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    $\begingroup$ The coadjoint orbits are not hyperKähler varieties. In fact, they are homogeneous and have positive Ricci curvature. $\endgroup$
    – Robert Bryant
    Commented Oct 29, 2014 at 13:56
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    $\begingroup$ Do you mean the projectivized coadjoint orbits? The coadjoint orbits are subvarieties of $\mathfrak{g}^*$, so not compact and so not projective. The orbits in $\mathbb{P}(\mathfrak{g}^*)$ are usually not called coadjoint orbits, I think. $\endgroup$
    – Ben McKay
    Commented Oct 30, 2014 at 15:35
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    $\begingroup$ @Hassan: Your question is about ${\it compact}$ Lie groups, in which case coadjoint orbits are ${\it not}$ hyperkaehler - consider for example $S^2$, a coadjoint orbit for $su(2)$. There is no need for Robert Bryant to revise his statement. The coadjoint orbits of a complex simple Lie algebra are another thing. $\endgroup$
    – Peter Dalakov
    Commented Oct 30, 2014 at 17:09
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    $\begingroup$ $G/P$ is birationally equivalent to its open Bruhat cell, a vector space. So birational invariants of $G/P$s are boring. $\endgroup$
    – Allen Knutson
    Commented Nov 3, 2014 at 14:21

1 Answer 1

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Since your manifold $X$ has positive Ricci curvature, the line bundles $K_X^m$ are all ${ negative}$ for $m\geq 1$, i.e. they admit a smooth Hermitian metric with negative curvature. By Kodaira Vanishing, we conclude that $H^0(X,K_X^m)=0$ for all $m\geq 1$, i.e. the Kodaira dimension is $-\infty$.

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  • $\begingroup$ Thanks YangMils@, you could you of also en.wikipedia.org/wiki/Borel%E2%80%93Weil%E2%80%93Bott_theorem , $\endgroup$
    – user21574
    Commented Oct 30, 2014 at 17:23
  • $\begingroup$ Indeed, $K_X = -2\rho$ so not dominant, under the usual indexing of line bundles by weights. $\endgroup$
    – Allen Knutson
    Commented Nov 2, 2014 at 10:49
  • $\begingroup$ Allen Knutson@, I need more detail, to undrestand $\endgroup$
    – user21574
    Commented Nov 2, 2014 at 14:17
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    $\begingroup$ A $G$-equivariant line bundle $\mathcal L$ on $G/B$ is uniquely determined by the $T$-weight on the line $\mathcal L|_{w_0 B/B}$ over the top point $w_0 B/B \in G/B$; if it's $\lambda$ (and dominant) then the space of sections is the irrep with high weight $\lambda$. Now, the tangent space $T_{B/B} G/B$ at the basepoint is $\mathfrak g/\mathfrak b$, with weights $\Delta_-$. Add those up and you get $-2\rho$. Now dualize (because we want the cotangent space, for $K_X$) and hit with $w_0$, obtaining $-2\rho$ again. In particular, when you add up all the negative roots, the result isn't dominant. $\endgroup$
    – Allen Knutson
    Commented Nov 3, 2014 at 14:30
  • $\begingroup$ Ohhh, thanks a lot, Now I undrestand that why Hirzebruch said that the first chern class of flag variaty is $2ρ$ $\endgroup$
    – user21574
    Commented Nov 3, 2014 at 14:35

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